The Physics Travel Guide

Coulomb's law

Prerequisites: Vector Calculus. Full path (3 pages)

F=keq1q2r2F= k_e \frac{q_1 q_2}{r^2}

Intuitive

Coulomb's law describes the force between two electric charges at rest. Charges of the same sign push each other apart, charges of opposite sign pull each other together. The force grows with the size of each charge and gets weaker with distance: at twice the distance it is four times weaker.

A useful picture for this "inverse square" behaviour is to imagine each charge sending out a fixed number of field lines in all directions. Far away, these lines are spread over the surface of a large sphere around the charge. The area of a sphere grows like the square of its radius, so the density of lines, and with it the strength of the force on a second charge, falls like one over the distance squared. The same counting is the content of Gauss' law.

The law has the same form as Newton's law of gravity, with charges in place of masses. There are two differences. Electric charge comes in two signs, so the force can attract or repel, and positive and negative charges in ordinary matter mostly cancel. And the electric force between elementary particles is enormously stronger than the gravitational one: between an electron and a proton it is about a thousand billion billion billion billion times larger. The Coulomb attraction between electrons and nuclei binds atoms and molecules.

Concrete

In vector form, the force on a point charge q1q_1 at position r⃗1\vec r_1 due to a point charge q2q_2 at r⃗2\vec r_2 is

F⃗12=keq1q2∣r⃗1−r⃗2∣2 r⃗1−r⃗2∣r⃗1−r⃗2∣,ke=14πε0≈8.99×109 N m2C2,\vec F_{12} = k_e \frac{q_1 q_2}{|\vec r_1 - \vec r_2|^2}\, \frac{\vec r_1 - \vec r_2}{|\vec r_1 - \vec r_2|}, \qquad k_e = \frac{1}{4\pi\varepsilon_0} \approx 8.99\times 10^{9}\,\frac{\text{N}\,\text{m}^2}{\text{C}^2},

in SI units, with ε0\varepsilon_0 the vacuum permittivity. For q1q2>0q_1 q_2 > 0 the force points away from q2q_2 (repulsion). It obeys F⃗21=−F⃗12\vec F_{21} = -\vec F_{12}. Forces from several charges add as vectors (superposition), so the force on a test charge qq can be written as F⃗=qE⃗\vec F = q\vec E with the electric field

E⃗(r⃗)=14πε0∑iqir⃗−r⃗i∣r⃗−r⃗i∣3=−∇ϕ,ϕ(r⃗)=14πε0∑iqi∣r⃗−r⃗i∣.\vec E(\vec r) = \frac{1}{4\pi\varepsilon_0}\sum_i q_i \frac{\vec r - \vec r_i}{|\vec r - \vec r_i|^3} = -\nabla\phi, \qquad \phi(\vec r) = \frac{1}{4\pi\varepsilon_0}\sum_i \frac{q_i}{|\vec r - \vec r_i|}.

From Gauss' law. For a point charge qq at the origin, symmetry says that E⃗\vec E is radial with a magnitude depending only on rr. Gauss' law ∮∂VE⃗⋅dA⃗=Qenc/ε0\oint_{\partial V}\vec E\cdot d\vec A = Q_{\text{enc}}/\varepsilon_0 applied to a sphere of radius rr gives

E(r) 4πr2=qε0⇒E(r)=q4πε0r2,E(r)\, 4\pi r^2 = \frac{q}{\varepsilon_0} \quad\Rightarrow\quad E(r) = \frac{q}{4\pi\varepsilon_0 r^2},

which is Coulomb's law. The factor 4πr24\pi r^2 is the area of the sphere, which is where the inverse square comes from.

Example. In the hydrogen atom, at the Bohr radius a0≈5.29×10−11 ma_0 \approx 5.29\times 10^{-11}\,\text{m}, the attraction between electron and proton is

F=e24πε0a02≈(8.99×109)(1.60×10−19)2(5.29×10−11)2 N≈8.2×10−8 N.F = \frac{e^2}{4\pi\varepsilon_0 a_0^2} \approx \frac{(8.99\times 10^9)(1.60\times 10^{-19})^2}{(5.29\times 10^{-11})^2}\,\text{N} \approx 8.2\times 10^{-8}\,\text{N}.

The ratio to the gravitational attraction, e2/(4πε0Gmemp)≈2.3×1039e^2/(4\pi\varepsilon_0 G m_e m_p) \approx 2.3\times 10^{39}, does not depend on the distance, since both forces fall like 1/r21/r^2.

Coulomb's law is the static limit of the inhomogenous Maxwell equations.

Abstract

Coulomb's law is the statement that the electrostatic potential of a point charge is, up to the factor q/ε0q/\varepsilon_0, the Green's function of the (negative) Laplacian in three dimensions. Electrostatics is governed by

∇⋅E⃗=ρε0,∇×E⃗=0⇒E⃗=−∇ϕ,−∇2ϕ=ρε0,\nabla\cdot\vec E = \frac{\rho}{\varepsilon_0}, \qquad \nabla\times\vec E = 0 \quad\Rightarrow\quad \vec E = -\nabla\phi, \quad -\nabla^2\phi = \frac{\rho}{\varepsilon_0},

and with ∇21∣x⃗∣=−4π δ3(x⃗)\nabla^2 \frac{1}{|\vec x|} = -4\pi\,\delta^3(\vec x) the solution that vanishes at infinity is

ϕ(x⃗)=14πε0∫d3x′ ρ(x⃗′)∣x⃗−x⃗′∣.\phi(\vec x) = \frac{1}{4\pi\varepsilon_0}\int d^3x'\, \frac{\rho(\vec x')}{|\vec x - \vec x'|}.

The same structure appears in momentum space. With the Fourier convention f(x⃗)=∫d3k(2π)3eik⃗⋅x⃗f~(k⃗)f(\vec x) = \int \frac{d^3k}{(2\pi)^3} e^{i\vec k\cdot\vec x}\tilde f(\vec k),

∫d3k(2π)3 eik⃗⋅x⃗k⃗2=14π∣x⃗∣,∫d3k(2π)3 eik⃗⋅x⃗k⃗2+m2=e−m∣x⃗∣4π∣x⃗∣.\int \frac{d^3k}{(2\pi)^3}\, \frac{e^{i\vec k\cdot\vec x}}{\vec k^2} = \frac{1}{4\pi|\vec x|}, \qquad \int \frac{d^3k}{(2\pi)^3}\, \frac{e^{i\vec k\cdot\vec x}}{\vec k^2 + m^2} = \frac{e^{-m|\vec x|}}{4\pi|\vec x|}.

In quantum field theory the potential between two static charges is read off from the exchange of a single virtual photon: in the non-relativistic limit the photon propagator reduces to 1/k⃗21/\vec k^2, and the first formula gives the Coulomb potential. In natural units with Heaviside-Lorentz conventions (ℏ=c=ε0=1\hbar = c = \varepsilon_0 = 1), the potential energy of an electron and a proton is V(r)=−e2/(4πr)=−α/rV(r) = -e^2/(4\pi r) = -\alpha/r with α≈1/137\alpha \approx 1/137. A photon mass mm would replace this by the Yukawa form of the second formula, so tests of the 1/r21/r^2 law are tests of the masslessness of the photon. Loop corrections (vacuum polarization) modify the potential at distances of order the electron Compton wavelength and below, where the effective coupling grows.

The power 1/r21/r^2 is tied to the dimension of space. In dd spatial dimensions the Green's function of the Laplacian behaves as r2−dr^{2-d} for d≥3d \ge 3 (and as ln⁡r\ln r for d=2d = 2), and the flux of the field through a sphere of area proportional to rd−1r^{d-1} fixes the force to fall like r1−dr^{1-d}. Covariantly, Coulomb's law is the static, point-source solution of the inhomogeneous Maxwell equations ∂μFμν=jν\partial_\mu F^{\mu\nu} = j^\nu (Heaviside-Lorentz, c=1c = 1, signature (+,−,−,−)(+,-,-,-), Fμν=∂μAν−∂νAμF^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu, so that the ν=0\nu = 0 component is ∇⋅E⃗=ρ\nabla\cdot\vec E = \rho) with jν=(q δ3(x⃗),0⃗)j^\nu = (q\,\delta^3(\vec x), \vec 0).

Definitions

Path to this page

  1. Calculus
  2. Linear Algebra
  3. Vector Calculus

Leads to: Maxwell Equations.